In YDSE, the distance of the slits from the screen is increased by $25 \%$ and the separation between the…

In YDSE, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If ' $\mathrm{W}$ ' represents the original fringe width, the new fringe width is
  1. $2 \mathrm{~W}$
  2. $2.5 \mathrm{~W}$
  3. 4W
  4. $1.5 \mathrm{~W}$

Solution

$\begin{aligned} & \mathrm{E}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=\frac{\mathrm{D}_2}{\mathrm{D}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2} \\ & \mathrm{D}_2=1.25 \mathrm{D}_1 \text { and } \mathrm{d}_2=\frac{\mathrm{d}_1}{2} \\ & \therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=1.25 \times 2=2.5 \\ & \therefore \mathrm{W}_2=2.5 \mathrm{~W}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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